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%I #38 Apr 09 2024 03:23:18
%S 2,162,2592,20000,101250,388962,1229312,3359232,8201250,18301250,
%T 37949472,74030112,137149922,243101250,414720000,684204032,1095962562,
%U 1710072162,2606420000,3889620000,5694792642,8194304162,11605565952,16200000000,22313281250,30356972802
%N a(n) = n^4*(n+1)^4/8.
%C A relation between fourth powers and the sum of fifth and seventh powers. See the first formula, which is from Beiler.
%D Albert H. Beiler, Recreations in the theory of numbers, New York, Dover, (2nd ed.) 1966, p. 161.
%H Temple Rice Hollcroft, <a href="http://dx.doi.org/10.1090/S0002-9904-1953-09755-5">On sums of powers of n consecutive integers</a>, Bulletin of the American Mathematical Society 59 (1953), nr. 6, p. 526 (574t).
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (9,-36,84,-126,126,-84,36,-9,1).
%F a(n) = 2*(Sum_{k=1..n} k)^4 = Sum_{k=1..n} (k^5 + k^7).
%F a(n) = 2*A059977(n-1).
%F a(n) = A000539(n) + A000541(n).
%F G.f.: -2*x*(1+72*x+603*x^2+1168*x^3+603*x^4+72*x^5+x^6) / (x-1)^9. - _R. J. Mathar_, Dec 13 2011
%F a(n) = 2*(A000217(n)^4). - _Zak Seidov_, Jan 21 2012
%F From _Amiram Eldar_, Apr 09 2024: (Start)
%F Sum_{n>=1} 1/a(n) = 8*Pi^4/45 + 80*Pi^2/3 - 280.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = 280 - 320*log(2) - 48*zeta(3). (End)
%p A202107:=n->(n^4)*(n+1)^4/8; seq(A202107(n), n=1..100); # _Wesley Ivan Hurt_, Nov 12 2013
%t Table[n^4 (n+1)^4/8, {n, 100}] (* _Wesley Ivan Hurt_, Nov 12 2013 *)
%Y Cf. A000217, A000539, A000541, A002117, A059977.
%K nonn,easy
%O 1,1
%A _Martin Renner_, Dec 11 2011
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